Power Supply Filter Anatomy

[GibsonGM]

I believe it is good to know the formulas, and how to get the answer on your own.  But you can see how easy it can be to make a simple error, so when you know the concept I think it is fine to use a calculator!   

I made some simple Visual Basic programs that do this work for me.  Some people also use Excel to help them work these problems out...

Yes, the forum is a great way to find an explanation that makes sense! 

[PRR]

> good to know the formulas

It is generally ample to know the reactance of a couple C-F pairs, and how to derive others.

100uFd is 100 Ohms at 20Hz. (15.9Hz, but never let "precision" get in the way of an easy answer.) Z of a C goes down when F goes up. So 100uFd is 17 Ohms at 120Hz usual US/Can ripple frequency.

47 Ohms and 100uFd is about 1/3rd ripple. 100 Ohms and 100uFd is about 1/6th ripple.

"Formulas" can often be simple ratios.

But is that enough? Too much? How sensitive is the pedal to ripple? How much ripple is audible with what signal strength?

And what nobody ever seems to say (and there is no calculator for) is: how much ripple is on that hunk of junk wart you are using? And if you respect your music, why use junk power?

[POTL]

> good to know the formulas

It is generally ample to know the reactance of a couple C-F pairs, and how to derive others.

100uFd is 100 Ohms at 20Hz. (15.9Hz, but never let "precision" get in the way of an easy answer.) Z of a C goes down when F goes up. So 100uFd is 17 Ohms at 120Hz usual US/Can ripple frequency.

47 Ohms and 100uFd is about 1/3rd ripple. 100 Ohms and 100uFd is about 1/6th ripple.

"Formulas" can often be simple ratios.

But is that enough? Too much? How sensitive is the pedal to ripple? How much ripple is audible with what signal strength?

And what nobody ever seems to say (and there is no calculator for) is: how much ripple is on that hunk of junk wart you are using? And if you respect your music, why use junk power?

Cool

Then i need 47uf/68R or 100uf/30R(33R) for my 50hz


odd one
favorite combination madbean site
47R and 100n(uf) gives the frequency of 33.9 hertz
for whom it is intended?
strange.
although in fact it does not matter

[merlinb]

Then i need 47uf/68R or 100uf/30R(33R) for my 50hz

Not really. You want as much filtering as possible, so you want the biggest capacitance and resistance that you can tolerate. The lower the cut-off frequency, the better the filtering.

[duck_arse]

.... which gives preimoschestva .....?

POTL, please, what is this word when translated to english?

[POTL]

.... which gives preimoschestva .....?

POTL, please, what is this word when translated to english?

This word is advantages преимущества=preimUschestva(I accidentally clicked "O" instead  "U") =advantages
which offer advantages? i did mean it

[duck_arse]

excellent, thank you.

[antonis]

The lower the cut-off frequency, the better the filtering.

The greater the cost, also.. 

Resistors are cheap and small & caps are expencive and large..

A cost & space saving guide should lead to BIG resistor and TINY capacitor but this should also result in greater voltage drop and worse ripple filtering..

For DIY projects the use of e.g 1000μF or 2200μF cap and 10R or 4R7 resistor isn't a big deal but for mass production it's surely a no-no..

[POTL]

I have finished testing the layouts and finally start building effects.
Prompt, there was a small question.
I chose a reverse polarity protection system using a 1N4001 diode in order not to lose voltage, but there is one nuance.
In some circuits, the resistor 47-100 ohms goes to the diode, in some after the protective diode, does it matter what?
I understand that the resistors do not have a polarity, but this resistor forms a filter together with the capacitors, if a diode is between the resistor and the capacitors, the filter will not work

[GibsonGM]

Is the diode in series, or in parallel?  I prefer the parallel method, with the diode opposed to normal current flow (cathode to "+").  This does not drop any voltage.   It is generally placed where the power enters the PCB. 

If the diode is in series, it appears likely that its low forward resistance when conducting would simply add to the 47-100 ohm resistor, no?  So it seems irrelevant...although you would lose approximately .5V if it is in series.

[POTL]

Hello - this is a parallel method, I also decided not to sacrifice the voltage;)

Especially the circuits that I build based on JFET and germanium transistors are quite sensitive to power.
I understand - I will install a resistor before the diode.
Thank you once again, as before the advice helped me)

[POTL]

By the way, how accurate is the information that when using the parallel method and the diode 1N4001 in the event of an incorrect polarity, the power supply will burn down?

[GibsonGM]

The way you have shown, the diode is not part of the circuit unless you have reversed polarity.  It is a very high resistance until it conducts, so you can ignore it.   

It DOES work in reverse polarity....if it does not burn out!  I believe a 1N400x diode can handle the power generated by 9V through it for a short time - it gets very hot...perhaps the way it is draws is a better way - if it were to conduct, it would have the 47R as a dropping resistor to prevent burnout.

Or, you can add a series resistor at the diode itself, 1K or other arbitrary value, to limit the current.  Again, it will not be involved unless you reverse polarity...this would ensure that the diode does not burn open, allowing the reversed polarity to pass into your circuit...

[POTL]

Thanks

[GibsonGM]

It does provide  reverse polarity protection!    

I am questioning the rating of the diode...will it open if it conducts 9V too long?  A battery may not be able to source sufficient current to destroy the diode on reversal, but as this is for a power supply, you may be able to cause it to open.   So the way it is placed may be a good way to do this - keep it in position with the 47R before it, to limit current on reversal.

I would use the circuit you posted; it looks good to me.


* any power reversal should be a short-term event...you should realize your mistake soon, maybe in less than 1 minute.  If the reversal continues, it is still possible to suffer component failure....example: the 47R resistor must take almost 2W if current is reversed.    A 1/4 W resistor will not survive very long should this occur.   It will get very hot, and probably become an open circuit, which will cut power to the rest of your circuit.    So you would have to replace it should this happen.    That is very much better than the reversed voltage going into the rest of your equipment!   

[POTL]

It does provide  reverse polarity protection!    

I am questioning the rating of the diode...will it open if it conducts 9V too long?  A battery may not be able to source sufficient current to destroy the diode on reversal, but as this is for a power supply, you may be able to cause it to open.   So the way it is placed may be a good way to do this - keep it in position with the 47R before it, to limit current on reversal.

I would use the circuit you posted; it looks good to me.

Sorry
I still have not raised my level of English and still communicate through Google translator (I think this is very noticeable)
Sometimes he translates sentences very strange, you have to read first in the original language, then in translation
With 3 attempts I realized what you wanted to say
It is still necessary to raise my level of English

[GibsonGM]

I understand, POTL, and I try to speak in a way you can understand.

[POTL]

I understand, POTL, and I try to speak in a way you can understand.

Thanks again!
Much grateful

[POTL]

And yes, I completely forgot
About the information that the power supply unit can burn out because of this method of protection
Is it true or not?)

[GibsonGM]

No.     The 47R can burn out.    It will create an open circuit.

The best solution is to use 47R  2W resistor.   

2W resistor will not burn out.